Engineering Economics

Er. Sushant Raj Giri B.E. (Industrial Engineering), MBA

Lecturer Department of Industrial Engineering

Contemporary Engineering Economics 3rd Edition Chan S Park

1

Lecture 5

2

Chapter 7 Present Worth Analysis

• Describing Project Cash Flows

• Initial Project Screening Method

• Present Worth Analysis • Variations of Present

Worth Analysis • Comparing Mutually

Exclusive Alternatives

3

Bank Loan vs. Investment Project

Bank Customer Loan

Repayment

Company Project

Investment

Return

� Bank Loan

� Investment Project

4

Describing Project Cash Flows Year (n)

Cash Inflows (Benefits)

Cash Outflows (Costs)

Net Cash Flows

0 0 $650,000 -$650,000

1 215,500 53,000 162,500

2 215,500 53,000 162,500

… … … …

8 215,500 53,000 162,500

5

Payback Period � Principle:

How fast can I recover my initial investment? � Method:

Based on cumulative cash flow (or accounting profit)

� Screening Guideline: If the payback period is less than or equal to some specified payback period, the project would be considered for further analysis.

� Weakness: Does not consider the time value of money

6

Example 7.3 Payback Period

N Cash Flow Cum. Flow

0 1 2 3 4 5 6

-$105,000+$20,000 $35,000 $45,000 $50,000 $50,000 $45,000 $35,000

-$85,000 -$50,000

-$5,000 $45,000 $95,000

$140,000 $175,000

Payback period should occurs somewhere between N = 2 and N = 3.

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-100,000

-50,000

0

50,000

100,000 150,000

0 1 2 3 4 5 6

Years (n)

3.2 years Payback period

$85,000

$15,000 $25,000

$35,000 $45,000 $45,000

$35,000

0 1 2 3 4 5 6

Years

Ann

ual c

ash

flow

Cum

ulat

ive

cash

flow

($)

8

Discounted Payback Period Calculation

Period Cash Flow Cost of Funds (15%)*

Cumulative Cash Flow

0 -$85,000 0 -$85,000

1 15,000 -$85,000(0.15)= -$12,750 -82,750

2 25,000 -$82,750(0.15)= -12,413 -70,163

3 35,000 -$70,163(0.15)= -10,524 -45,687

4 45,000 -$45,687(0.15)=-6,853 -7,540

5 45,000 -$7,540(0.15)= -1,131 36,329

6 35,000 $36,329(0.15)= 5,449 76,778

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Net Present Worth Measure � Principle: Compute the equivalent net surplus at n = 0 for a given interest rate of i. � Decision Rule: Accept the project if the net surplus is positive.

2 3 4 5 0 1

Inflow

Outflow

0 PW(i) inflow

PW(i) outflow

Net surplus

PW(i) > 0

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Example 7.5 - Tiger Machine Tool Company

PW P F P FP F

PWPW

( $24, ( / , ) $27, ( / , )$55, ( / , )$78,

( $75,( $78, $75,

$3, ,

15%) 400 15%,1 340 15%,2760 15%,3553

15%) 00015%) 553 000

553 0

inflow

outflow

Accept

= ++=== −= >

$75,000

$24,400 $27,340 $55,760

0 1 2 3

outflow

inflow

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Present Worth Amounts at Varying Interest Rates

i (%) PW(i) i(%) PW(i) 0 $32,500 20 -$3,412 2 27,743 22 -5,924 4 23,309 24 -8,296 6 19,169 26 -10,539 8 15,296 28 -12,662 10 11,670 30 -14,673 12 8,270 32 -16,580 14 5,077 34 -18,360 16 2,076 36 -20,110

17.45* 0 38 -21,745 18 -751 40 -23,302

*Break even interest rate

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-30

-20

-10

0

10

20

30

40

0 5 10 15 20 25 30 35 40

PW (i

) ($

thou

sand

s)

i = MARR (%)

$3553 17.45%

Break even interest rate (or rate of return)

Accept Reject

Present Worth Profile

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Future Worth Criterion

• Given: Cash flows and MARR (i)

• Find: The net equivalent worth at the end of project life

$75,000

$24,400 $27,340 $55,760

0 1 2 3

Project life

14

FW F P F PF P

FW F P

FW

( $24, ( / , ) $27, ( / , )$55, ( / , )$119,

( $75, ( / , )$114,

( $119, $114,$5, ,

15%) 400 15%,2 340 15%,1760 15%,0470

15%) 000 15%,3066

15%) 470 066404 0

inflow

outflow

Accept

= +

+

=

=

=

= −

= >

Future Worth Criterion

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Project Balance Concept N 0 1 2 3 Beginning

Balance

Interest

Payment

Project Balance

-$75,000

-$75,000

-$75,000

-$11,250

+$24,400

-$61,850

-$61,850

-$9,278

+$27,340

-$43,788

-$43,788

-$6,568

+$55,760

+$5,404

Net future worth, FW(15%)

PW(15%) = $5,404 (P/F, 15%, 3) = $3,553

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Project Balance Diagram

60,000

40,000

20,000

0

-20,000

-40,000

-60,000

-80,000

-100,000

-120,000

0 1 2 3

-$75,000 -$61,850

-$43,788

$5,404

Year(n)

Terminal project balance (net future worth, or

project surplus)

Discounted payback period

Proj

ect b

alan

ce ($

)

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Capitalized Equivalent Worth

� Principle: PW for a project with an annual receipt of A over infinite service life �Equation:

CE(i) = A(P/A, i, ) = A/i A

0

P = CE(i)

∞

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Given: i = 10%, N = Find: P or CE (10%)

10

$1,000 $2,000

P = CE (10%) = ?

0

∞

CE P F( $1,.

$1,.

( / , )

$10, ( . )$13,

10%) 0000 10

0000 10

10%,10

000 1 0 3855855

= +

= +

=

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Example 7.9 Mr. Bracewell’s Investment Problem

• Built a hydroelectric plant using his personal savings of $800,000

• Power generating capacity of 6 million kwhs

• Estimated annual power sales after taxes - $120,000

• Expected service life of 50 years � Was Bracewell's $800,000 investment a wise one? � How long does he have to wait to recover his initial investment, and will he ever make a profit?

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Mr. Brcewell’s Hydro Project

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Equivalent Worth at Plant Operation • Equivalent lump sum investment

V1 = $50K(F/P, 8%, 9) + $50K(F/P, 8%, 8) + . . . + $100K(F/P, 8%, 1) + $60K

= $1,101K

• Equivalent lump sum benefits

V2 = $120(P/A, 8%, 50) = $1,468K

• Equivalent net worth FW(8%) = V1 - V2

= $367K > 0, Good Investment

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With an Infinite Project Life • Equivalent lump sum investment

V1 = $50K(F/P, 8%, 9) + $50K(F/P, 8%, 8) + . . . + $100K(F/P, 8%, 1) + $60K

= $1,101K

• Equivalent lump sum benefits assuming N = V2 = $120(P/A, 8%, ∞ )

= $120/0.08 = $1,500K

• Equivalent net worth FW(8%) = V1 - V2

= $399K > 0 Difference = $32,000

∞

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Problem 7.27 - Bridge Construction

� Construction cost = $2,000,000 � Annual Maintenance cost = $50,000 � Renovation cost = $500,000 every 15 years � Planning horizon = infinite period � Interest rate = 5%

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$500,000 $500,000 $500,000 $500,000

$2,000,000

$50,000

0 15 30 45 60

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Solution: • Construction Cost

P1 = $2,000,000 • Maintenance Costs

P2 = $50,000/0.05 = $1,000,000 • Renovation Costs

P3 = $500,000(P/F, 5%, 15) + $500,000(P/F, 5%, 30) + $500,000(P/F, 5%, 45) + $500,000(P/F, 5%, 60) . = {$500,000(A/F, 5%, 15)}/0.05 = $463,423

• Total Present Worth P = P1 + P2 + P3 = $3,463,423

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Alternate way to calculate P3 • Concept: Find the effective interest rate per payment period

• Effective interest rate for a 15-year cycle i = (1 + 0.05)15 - 1 = 107.893% • Capitalized equivalent worth

P3 = $500,000/1.07893 = $463,423

15 30 45 60 0

$500,000 $500,000 $500,000 $500,000

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Comparing Mutually Exclusive Projects

� Mutually Exclusive Projects � Alternative vs. Project � Do-Nothing Alternative

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�Revenue Projects

Projects whose revenues depend on the choice of alternatives

�Service Projects

Projects whose revenues do not depend on the choice of alternative

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�Analysis Period The time span over which the economic effects of an investment will be evaluated (study period or planning horizon).

�Required Service Period The time span over which the service of an equipment (or investment) will be needed.

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Comparing Mutually Exclusive Projects

• Principle: Projects must be compared over an equal time span. • Rule of Thumb: If the required service period is given, the analysis period should be the same as the required service period.

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Finite

Required service period

Infinite

Analysis = Required period service period

Project repeatability likely

Project repeatability unlikely

Analysis period equals

project lives

Analysis period is longest of project life in the group

Analysis period is lowest common multiple

of project lives

Analysis period equals one of

the project lives

Analysis period is shorter than

project lives

Analysis period is longer than project lives

Case 1

Case 2

Case 3

Case 4

How to Choose An Analysis Period

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Case 1: Analysis Period Equals Project Lives

Compute the PW for each project over its life

$450 $600

$500 $1,400 $2,075

$2,110

0

$1,000 $4,000 A B

PW (10%) = $283 PW (10%) = $579

A

B

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$1,000

$450 $600

$500

Project A

$1,000

$600 $500 $450

$3,000

3,993

$4,000

$1,400

$2,075

$2,110

Project B Modified Project A

Comparing projects requiring different levels of investment – Assume that the unused funds will be invested at MARR.

PW(10%)A = $283 PW(10%)B = $579

This portion of investment will earn 10% return on investment.

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Case 2: Analysis Period Shorter than Project Lives

• Estimate the salvage value at the end of required service period.

• Compute the PW for each project over the required service period.

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Comparison of unequal-lived service projects when the required service period is shorter than the

individual project life

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Case 3: Analysis Period Longer than Project Lives

• Come up with replacement projects that match or exceed the required service period.

• Compute the PW for each project over the required service period.

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Comparison for Service Projects with Unequal Lives when the required service period is longer than the

individual project life

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Case 4: Analysis Period is Not Specified

• Project Repeatability Unlikely Use common service (revenue) period.

• Project Repeatability Likely Use the lowest common multiple of project lives.

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Project Repeatability Unlikely

Assume no revenues

PW(15%)drill = $2,208,470

PW(15%)lease = $2,180,210

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Project Repeatability Likely

PW(15%)A=-$53,657

PW(15%)B=-$48,534

Model A: 3 Years Model B: 4 years LCM (3,4) = 12 years

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Summary • Present worth is an equivalence method of analysis in

which a project’s cash flows are discounted to a lump sum amount at present time.

• The MARR or minimum attractive rate of return is the interest rate at which a firm can always earn or borrow money.

• MARR is generally dictated by management and is the rate at which NPW analysis should be conducted.

• Two measures of investment, the net future worth and the capitalized equivalent worth, are variations to the NPW criterion.

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• The term mutually exclusive means that, when one of several alternatives that meet the same need is selected, the others will be rejected.

• Revenue projects are those for which the income generated depends on the choice of project.

• Service projects are those for which income remains the same, regardless of which project is selected.

• The analysis period (study period) is the time span over which the economic effects of an investment will be evaluated.

• The required service period is the time span over which the service of an equipment (or investment) will be needed.

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• The analysis period should be chosen to cover the

required service period. • When not specified by management or company

policy, the analysis period to use in a comparison of mutually exclusive projects may be chosen by an individual analyst.

End of Lecture 5